Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a question related to an earlier question of mine:

I've been reading about topological invariants. Some of them are defined in terms of quadratic forms.

My current understanding is: we can turn a topological space $X$ into a module by taking any of its homologies and once we have a module we define a bilinear form on $H_i(X)$. For example, if we are taking the first homology we can define an bilinear form by defining $x \cdot y$ to be the intersection number of two representatives $x,y$ where $[x], [y] \in H_1 (X)$. Then this gives us a quadratic form $Q(x) = \frac{1}{2} x \cdot x \mod 2$. We can use this if we want to show that two topological spaces are not homeomorphic as follows: pick a symplectic basis $a_i, b_i$ for the space (we know that such a basis always exists). Then we define $A(Q) = \sum_i Q(a_i) Q(b_i)$. If this evaluates to $0$ in one space and to $1$ in the other we know that the two spaces are not homeomorphic.


Question 1: Is my current understanding correct? Or am I missing anything?

Question 2: I understand how to define a bilinear form if I take the first homology. But this might not lead anywhere because even though the two spaces might be non-homeomorphic, their first homology groups might still coincide. So I might want to take a higher homology. How do I define a bilinear form for the $k$-th homology and what is the geometric meaning?

Question 3: I found this Wikipedia article about intersection forms. Although I am quite sure that the "intersection form" is another bilinear form just like the intersection number for the first homology, I'm confused about why I could only find additional information about $4$-manifolds. What is special about $4$-manifolds? Am I wrong in assuming that I can endow any topological space with a bilinear form?

Question 4: We do this over $\mathbb Z / 2$ for convenience, right? Because it lets us ignore orientation. But we could also consider bilinear forms over any other field, is this correct?

Edit

Question 5: Poincaré duality gives us a way to define an intersection pairing without much nasty fiddling if the topological space $X$ is a closed oriented manifold. But what are the absolute minimal requirements on $X$ in order to be able to define an intersection form on it?

Thanks for your help.

share|improve this question
1  
There is at least a way to generalise the 4-dimensional intersection form to a general billinear form on $2n$-dimensional connected oriented manifolds –  Juan S Aug 24 '12 at 2:24
    
Dear @JuanS, thank you for your comment! To $n$-manifold won't work? Why does it have to be $2n$? –  Matt N. Aug 24 '12 at 5:50
2  
@Matt: the cup product gives a map $H^i \times H^j \to H^{i+j}$. If you want $i = j$ then you get a map $H^i \times H^i \to H^{2i}$. You want $H^{2i}$ to be the top cohomology so you can use Poincaré duality to identify it with $H_0$ and this requires that the manifold be $2i$-dimensional. –  Qiaochu Yuan Aug 24 '12 at 6:07
1  
@Matt: also, it's inaccurate to call these things inner products. An inner product is always positive-definite. –  Qiaochu Yuan Aug 24 '12 at 6:07
    
Dear @QiaochuYuan, thank you for your helpful comments! I corrected the question. –  Matt N. Aug 24 '12 at 6:10
show 1 more comment

1 Answer

up vote 4 down vote accepted

Q1: no. The intersection pairing is not defined on general topological spaces.

Q2: Your space will need some special structure and your question is still too general to answer. Perhaps read-up on Poincare duality?

Q3: An intersection form is a special bilinear form induced via Poincare duality. You can think of it in terms of (oriented) transverse intersections of representative cycles.

Q4: Yes, but generally you'll need your manifold to be oriented to pull this off.

share|improve this answer
    
Re Q2: Sorry could you give me an example of what sort of special additional structure I need? Then I can ask a more precise question in a new thread. Thanks. –  Matt N. Aug 24 '12 at 12:49
    
Re Q1: Then what do I need to require in order to define the intersection pairing? (I assume intersection pairing and intersection number are the same thing) –  Matt N. Aug 24 '12 at 12:54
    
I did suggest an example -- look up Poincare duality. That's where intersection pairings are defined. I suppose it's hard for me to understand what you have in mind if you don't seem to be using a definition for a term you're talking about. –  Ryan Budney Aug 25 '12 at 4:16
1  
What properties would you like such an "intersection form" to have? –  Ryan Budney Aug 28 '12 at 16:20
1  
With such a weak collection of conditions, bilinear forms are not even partially ordered. So you've got no hope of a "minimal condition" for one to exist. You need to give the pairing some kind of non-trivial character / flavour. Have you seen the torsion linking form for example? There's lots of bilinear forms you can put on homology, depending on the kind of space you have. –  Ryan Budney Aug 31 '12 at 20:46
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.